I have a document with multiple information. What I want is to build a Notepad++ Regex replace function, that finds the following lines in the document and replaces the blank spaces between the "" with an underline (_).
Example:
The line is:
&LOG Part: "NAME TEST.zip"
The result should be:
&LOG Part: "NAME_TEST.zip"
The perfect solution would be that the regex finds the &LOG Part: "NAME TEST.zip" lines and replaces the blank space with an underline.
What I have tried for now is this expression to find the text between the " ":
\"[^"]*\"
It should do it, but I don't know which expression to use to replace the blank spaces with an underline.
Anyone could help with a solution?
Thanks!
The \"[^"]*\" will only match whole substrings from " up to another closest " without matching individual spaces you want to replace.
Since Notepad++ does not support infinite width lookbehind, the only possible solution is using the \G - based regex to set the boundaries and use multiple matching (this one will replace consecutive spaces with 1 _):
(?:"|(?!^)\G)\K([^ "]*) +(?=[^"]*")
Or (if each space should be replaced with an underscore):
(?:"|(?!^)\G)\K([^ "]*) (?=[^"]*")
And replace with $1_. If you need to restrict to replacing inside &LOG Part only, just add it to the beginning:
(?:&LOG Part:\s*"|(?!^)\G)\K([^ "]*) (?=[^"]*")
A human-readable explanation of the regex:
(?:"|(?!^)\G)\K - Find a ", or, with each subsequent successful match, the end of the previous successful match position, and omit all the text in the buffer (thanks to \K)
([^ "]*) - (Group 1, accessed with$1from the replacement pattern) 0+ characters other than a space and"`
+ - one or more literal spaces (replace with \h to match all horizontal whitespace, or \s to match any whitespace)
(?=[^"]*") - check if there is a double quote ahead of the current position
Related
I'm working in notepad++, and using its find-replace dialog box.
NP++ documentation states: Notepad++ regular expressions use the Boost regular expression library v1.70, which is based on PCRE (Perl Compatible Regular Expression) syntax. ref: https://npp-user-manual.org/docs/searching
What I'm trying to do should be simple, but I'm a regex novice, and after 2-3 hrs of web searches and playing with online regex testers, I give up.
I want to replace all single quotes ' with double quote " , but if and only if the ' is to the RIGHT of one or more #, ie inside a python comment.
For example,
list1 = ['apple','banana','pear'] # All 'single quotes' to LEFT of # remained unchanged.
list2 = ['tomato','carrot'] # All 'single quotes' to RIGHT of one or more # are replaced
# # with "double quotes", like this.
The np++ file is over 800 lines, manual replacement would be tedious & error prone. Advice appreciated.
This regex should do what you want:
(^[^#]*#|(?<!^)\G)[^'\n]*\K'
It looks for a ' which is preceded by either
^[^#]*# : start of line and some number of non-# characters followed by a #; or
(?<!^)\G : the start of line or the end of the previous match (\G), with a negative lookbehind for start of line (?<!^), meaning that it only matches at the end of the previous match
and then some number of non ' or newline (to prevent the match wrapping around the end of the previous line) characters [^'\n]*.
We then use \K to reset the match, so that everything before that is discarded from the match, and the regex only matches the '.
That can then be replaced with ".
Demo on regex101
Update
You can avoid matching apostrophes within words by only matching ones that are either preceded or followed by a non-word character:
(^[^#]*#|(?<!^)\G)[^'\n]*\K('(?=\W)|(?<=\W)')
Demo on regex101
Update 2
You can also deal with the case where there are # characters in strings by qualifying the first part of the regex with the requirement for there to be matched pairs of quotes beforehand:
(?:^[^'#]*(?:'[^']*'[^#']*)*[^'#]*#|(?<!^)\G)[^'\n]*\K(?:'(?=\W)|(?<=\W)')
Demo on regex101
I want to replace character - using regular expression in my text so it would work like this:
Original text: abcd-efg-hijk-lmno
Text after replacing: abcd-efg-hijk/lmno
As you can see I want to replace character - starting from the end just one time with character /.
Thanks in advance for any tips
Find what: -([^-]*)$
Replace with: /$1
Search Mode: Regular Expression
Explanation:
- : a dash
([^-]*$) : text with no dash,
zero or more times,
to the end of the line,
put in the $1 variable
/$1 : literal "/", contents of $1
Good resource: http://www.grymoire.com/Unix/Regular.html
To replace characters in Notepad++, you can open the Replace window using Ctrl+H, or under the "Search" menu. Once open, enter the following regular expression:
(.{4}-.{3}-.{4})(-)(.{4})
This will find:
a group of four characters (the "." being any character, the "{4}" being the quantity),
a dash,
a group of three characters,
another dash,
a group of four characters,
again another dash,
then a group of four characters.
The parentheses group this search into captured groups, which we will use for the replacement part. See https://www.regular-expressions.info/brackets.html for more info.
If you want to restrict the search to lowercase letters as in your example, you would replace the "." with "[a-z]", or for upper and lower "[a-z,A-Z]".
Now for the replacement. The groups from earlier are referenced by the dollar sign then the number, e.g. $1 would be the first. So we will replace the characters found with the first group ($1), disregard the second group containing the dash and insert the "/" instead, then include the third group ($3):
$1/$3
The settings in the replace window need to have "Regular expression" and "Wrap around" checked, and ". matches newline" unchecked.
You can then click Replace all to replace all occurrences, or go through using Replace individually.
Since the beginning and end of line characters are not included, you can find multiple occurrences of this pattern on a single line.
Note: This answer follows the same procedure as Toto's, however uses a different regular expression.
Ctrl+H
Find what: ^(.+)-([^-]+)$
Replace with: $1/$2
check Wrap around
check Regular expression
DO NOT CHECK . matches newline
Replace all
Explanation:
^ : begining of line
(.+) : 1 or more any character, catch in group 1
- : a dash
([^-]+) : 1 or more any character but dash, catch in group 2
$ : end of line
In my script, I'm in passing a markdown file and using sed, I'm trying to find lines that do not have one or more # and are not empty lines and then surround those lines with <p></p> tags
My reasoning:
^[^#]+ At beginning of line, find lines that do not begin with 1 or more #
.\+ Then find lines that contain one or more character (aka not empty lines)
Then replace the matched line with <p>\1</p>, where \1 represents the matched line.
However, I'm getting "\1 not defined in the RE". Is my reasoning above correct and how do I fix this error?
BODY=$(sed -E 's/^[^#]+.\+/<p>\1</p>/g' "$1")
Backslash followed by a number is replaced with the match for the Nth capture group in the regexp, but your regexp has no capture groups.
If you want to replace the entire match, use &:
BODY=$(sed -E 's%^[^#].*%<p>&</p>%' "$1")
You don't need to use .+ to find non-empty lines -- the fact that it has a character at the beginning that doesn't match # means it's not empty. And you don't need + after [^#] -- all you care is that the first character isn't #. You also don't need the g modifier when the regexp matches the entire line -- that's only needed to replace multiple matches per line.
And since your replacement string contains /, you need to either escape it or change the delimiter to some other character.
I'm trying to replace whitespaces with underscores in certain parts of my html-document with Notepad++.
I can identify the area to search for the whitespaces in the following way:
-Begins with: src="video/
-Ends with: mp4
For example I might have a line like this:
<video class="play" src="video/my file name with empty spaces.mp4">
and I would like to change it to be like this:
<video class="play" src="video/my_file_name_with_empty_spaces.mp4">
Tested in N++
Search: (?:src="video|(?<!^)\G)(?:(?!mp4).)*?\K\s+
Replace: _
On the demo, see the substitutions at the bottom.
Explanation
(?:src="video|(?<!^)\G) matches the delimiter src="video, or \G the position following the previous match as long as it is not at the beginning of the string (?<!^) where \G can also match
(?:(?!mp4).) matches one character that is not followed by mp4
*? lazily matches such characters, up to...
\s a space character (our match which we replace with _)
before the space, the \K tells the engine to drop what was matched so far from the final match it returns
I have a large file with content inside every bracket. This is not at the beginning of the line.
1. Atmos-phere (7800)
2. Atmospheric composition (90100)
3.Air quality (10110)
4. Atmospheric chemistry and composition (889s120)
5.Atmospheric particulates (10678130)
I need to do the following
Replace the entire content, get rid of line numbers
1.Atmosphere (10000) to plain Atmosphere
Delete the line numbers as well
1.Atmosphere (10000) to plain Atmosphere
make it a hyperlink
1.Atmosphere (10000) to plain linky study
[I added/Edit] Extract the words into a new file, where we get a simple list of key words. Can you also please explain the numbers in replace the \1\2, and escape on some characters
Each set of key words is a new line
Atmospheric
Atmospheric composition
Air quality
Each set is a on one line separated by one space and commas
Atmospheric, Atmospheric composition, Air quality
I tried find with regex like so, \(*\) it finds the brackets, but dont know how to replace this, and where to put the replace, and what variable holds the replacement value.
Here is mine exression for notepad ([0-9(). ]*)(.*)(\s\()(.*)
You need split your search in groups
([0-9. ]*) numbers, spaces and dots combination in 0 or more times
(.*) everything till next expression
(\s\() space and opening parenthesis
(.*) everything else
In replace box - for practicing if you place
\1\2\3\4 this do nothing :) just print all groups from above from 1.1 to 1.4
\2 this way you get only 1.2 group
new_thing\2new_thing adds your text before and after group
<a href=blah.com/\2.html>linky study</a> so now your text is added - spaces between words can be problematic when creating link - so another expression need to be made to replace all spaces in link to i.e. _
If you need add backslash as text (or other special sign used by regex) it must be escaped so you put \\ for backslash or \$ for dolar sign
Want more tune - <a href=blah.com/\2.html>\2</a> add again 1.2 group - or use whichever you want
On the screenshot you can see how I use it (I had found and replaced one line)
Ok and then we have case 4.2 with colon at the end so simply add colon after extracted section:
change replace from \2 to \2,
Now you need join it so simplest way is to Edit->Line Operations->Join Lines
but if you want to be real pro switch to Extended mode (just above Regular expression mode in Replace window) and Find \r\n and replace with space.
Removing line endings can differ in some cases but this is another story - for now I assume that you using windows since Notepad++ is windows tool and line endings are in windows style :)
The following regex should do the job: \d+\.\s*(.*?)\s*\(.*?\).
And the replacement: <a href=example.com\\\1.htm>\1</a>.
Explanation:
\d+ : Match a digit 0 or more times.
\. : Match a dot.
\s* : Match spaces 0 or more times.
(.*?) : Group and match everything until ( found.
\s* : Match spaces 0 or more times.
\(.*?\) : Match parenthesis and what's between it.
The replacement part is simple since \1 is referring to the matching group.
Online demo.
Try replacing ^\d+\.(.*) \(\w+\)$ with <a href=blah.com\\\1.htm>linky study</a>.
The ^\d+. removes the leading number and dot. The (.*) collects the words. Then there is a single space. The \(\w+\)$ matches the final number in brackets.
Update for the added Q4.
Regular expressions capture things written between round brackets ( and ). Brackets that are to be found in the text being searched must be escaped as \( and \). In the replacement expression the \1 and \2 etc are replaced by the corresponding capture expression. So a search expression such as Z(\d+)X([aeiou]+)Y might match Z29XeieiY then the replacement expression P\2Q\1R would insert PeieiQ29R. In the search at the top of this answer there is one capture, the (.) captures or collects the words and then the \1 inserts the captured words into the replacement text.